The Impossibility of Flying Heavy Aircraft Without Training

OK, I'll buy this. I should remind everyone that I am still
talking about an infinite wing, having no induced drag.

I claim that my statement is generally true, and does not require an
infinite wing, just an infinite air field (as per the example I'm
developing).

However, air will still have to be accelerated downwards to
keep the airplane from succumbing to gravity.


No it won't. Now we have no acceleration of the plane
upwards. It's true that the wing must still throw air down,
but now it's catching the air that is rising ahead of the
wing, and throwing it down at a speed that exactly matches
the rising speed of the air. Thus, the net downward speed
of the air is zero and the wing leaves undisturbed air
behind it..

Yes, it will.

The rising air in front of the wing doesn't rise by magic. It doesn't
rise before the wing does its thing (though it may rise before the wing
gets there); it rises =because= the wing is doing its thing, which is
throwing the air down (where it pushes the other air aside, and that
other air has to go somewhere).

Think about it. There's no earth, just this "magic plane gravity" which
pushes the airplane down. Something has to push up on the plane to keep
it in the air. The air can't push up (the way the runway can) because
the air is fluid - it smooshes out of the way and lets the plane
through. The only way the air can push up is by being thrown down -
accelerated at a rate that matches the acceleration due to "magic plane
gravity".

A wing not only keeps the plane away from the earth, it keeps the earth
away from the plane. If you could measure the total forces on the earth
due to everything on top of it (essentially making the earth a giant
bathroom scale), the reading would not change when an airplane takes
off. Even though the plane is not touching the earth, it is throwing
air down at the earth, and that impact registers as weight.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

'Dissipation' is flow. If you include that dissipation into your
sum, there is no net flow. Otherwise, as stated above, the
ambient pressure at ground level would steadily increase and,
as you note below, the pressure higher up woudl steadily drop.

But the ambient pressure at ground level =does= increase, by an amount
equal to the weight of the airplane (divided by the surface area of the
earth). This remains true for as long as the airplane is being
"supported" by the air.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

wrote:

In the real world there are many photographs of huge canyons carved in
layers of cloud and smoke as airplanes fly over them,

Cool! Got any links to some? How about pictures of airplanes
flying just below the ceiling?


The most famous photo is one that you can see on this site he
http://adamone.rchomepage.com/index4.htm
He has the caption wrong, though. The Citation never flew through the
cloud, only over it.

Pictures of airplanes flying just below the ceiling are an interesting
idea, but I have not seen any. Usually the ceiling is fairly ill
defined and ragged anyway.

Jose wrote:

Occom's Razor, Jose.


Yes, but Occam's razor needs to work with theories that explain the
facts. Fact is, the down on an airplane =is= laterally separated from
the up, and airplanes fly only when they move forward.

If Occam's razor were so simple, we wouldn't have million dollar grants.

Jose

OK, Jose (just had to say that one) explain delta wings.

http://ernest.isa-geek.org

--
This is by far the hardest lesson about freedom. It goes against
instinct, and morality, to just sit back and watch people make
mistakes. We want to help them, which means control them and their
decisions, but in doing so we actually hurt them (and ourselves)."

Jose wrote:
I think you are refering to Newton's third law, often stated as: "For
every action there is an equal and opposite reaction."

Yes.

For an aircraft in level flight, the upwards acceleration due to lift
is counterbalanced by the downward acceleration due to gravity.
This satisfies Newton's third law.

Yes.

For a wing in level flight, the vertical component of momentum is
zero.

No.

Please show us your arithmetic. Suppose a 1500 lb airplane is
flying horizontally at 120 mph at 5000 feet above MSL. What
are the vertical and horizontal components of the momentum
of that aircraft?


That is, on a microscopic scale, no. The wing is constantly
freefalling, then being bounced back up by impact with air molecules.
Averaged over all the molecules, yes, the net is zero (the wing flies)
but on a microscopic scale, the wing is in constant brownian motion.
This implies momentum transfer, and following the momentum on a
microscopic scale is instructive.


OK, show us your arithmetic.

The wing imparts
as much upward momentum to the air as it does downward momentum.

This is where I disagree. Upward momentum gets imparted, but not
(directly) by the wing. Rather, it is imparted by the ground, mediated
through other air molecules.

The ground is stationary. How does the stationary ground impart
momentum to anything?

Of course this wouldn't happen if the wing
didn't pass through and throw the air down to begin with, but the ground
is what ultimately imparts the upwards momentum.

The pressure differential through the wing, from bottom to top,
integrated
over the wing area, provides an upward force for a wing in level
flight.

That's the shortcut. Where does this pressure differential come from -

Bernouli effect.

that is the question.

The downwash behind the aircraft, which is counterbalanced by a more
diffuse upwash around it, is real but not relevent to the issue of
lift.

I disagree here too. It's important in seeing the entire picture.

Well, yes it is part of the entire picture. Its just not relevent to
the
issue of lift, which is only part of the picture.

--

FF

Dan wrote in news:L_1Nf.55333$Ug4.44233@dukeread12:

george wrote:
Richard Lamb wrote:
Al wrote:
Remember, Time flies like an arrow...
Fruit Flies Like a Banana...

Do all fruit flies have bannana ratings ?
:-)


Only visual banana rules in most cases. I heard the IBR are a tad
tough.

Dan, U.S. Air Force, retired


The scarcity of tiny instrument training hoods or banana simulators makes
it even more difficult.

OK, show us your arithmetic.

First, do you agree that air is made of individual molecules separated
by a lot of space compared to the size of the molecules themselves?
Then do you accept that a wing is in freefall during all the (very brief
but very numerous) time in between molecular collisions? (If not, what
holds it up when it is not in contact with any air molecules?)

If so, then during the time it is in freefall, it acquires a downward
velocity. Small, no doubt, but nonzero. The next molecular impact
pushes it back up. On the average they will sum to a net zero vertical
motion. Is this the arithmetic you want to see?

The ground is stationary. How does the stationary ground impart
momentum to anything?

The ground is not stationary. Like the wing, the ground is jiggling
around in brownian motion. Such motion is greatly overwhelmed in
quantity by other things, but it is nonzero. Gravity pulls the ground
towards the airplane just as strongly as it pulls the airplane towards
the ground. This is the same effect as the one that gives high tides on
the side of the earth that is away from the moon.

Where does this pressure differential come from -

Bernouli effect.

That's the shortcut. Where does the Bernoulli effect come from - on a
molecular level? That's what I'm addressing. The Bernoulli effect is a
shortcut for doing the calculation in bulk (where it makes the most
sense if you want a numerical answer) but it all comes from molecular
collisions.

Its just not relevent to the
issue of lift, which is only part of the picture.

We disagree here. Both explanations are true as far as they go, but it
is important to see just how far they go (or don't go). The Beruoulli
effect does not explain, for example, how the earth ultimately supports
the aircraft, nor how the upwash starts (for example, suppose there were
a vertical column of vacuum separated from the air by a very strong
piece of cellophane. A wing travels through the vacuum and penetrates
this cellophane. The air behind the cellophane does not rise up to meet
the wing - it has no idea there's a wing coming. Once the wing has
entered the air, that rising motion will start, but why?

That's the question to which I am applying my molecular model.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

Richard Lamb wrote:

Sez one old fart to another???

Al wrote:

Richard you give away your age. Only an "Old Fart" would know that.

Al


"Richard Lamb" wrote in message
ink.net...

Al wrote:

Remember, Time flies like an arrow...
Fruit Flies Like a Banana...


Grouch Marx!




Well _I_ really do know how to spell it.
and now, my spell checker does too...

Jose wrote:
fredfighter wrote:

Please show us your arithmetic. Suppose a 1500 lb airplane is
flying horizontally at 120 mph at 5000 feet above MSL. What
are the vertical and horizontal components of the momentum
of that aircraft?

That is, on a microscopic scale, no. The wing is constantly
freefalling, then being bounced back up by impact with air molecules.
Averaged over all the molecules, yes, the net is zero (the wing flies)
but on a microscopic scale, the wing is in constant brownian motion.
This implies momentum transfer, and following the momentum on a
microscopic scale is instructive.

OK, show us your arithmetic.

First, do you agree that air is made of individual molecules separated
by a lot of space compared to the size of the molecules themselves?

Yes.

Then do you accept that a wing is in freefall during all the (very brief
but very numerous) time in between molecular collisions? (If not, what
holds it up when it is not in contact with any air molecules?)

Yes.


If so, then during the time it is in freefall, it acquires a downward
velocity. Small, no doubt, but nonzero.

Sometimes it does, sometimes it does not. I'll allow as the vertical
component of velocity decreases during that time, for a positive up
coordinate system and a plane in (macroscopic) level flight.

Do you agree that in each collision momentum is transferred to the
air molecule that is equal and opposite to the momentum transferred
to the wing?

The next molecular impact
pushes it back up. On the average they will sum to a net zero vertical
motion. Is this the arithmetic you want to see?

No, I want you to calculate the horizontal and vertical componenets
of momentum for the example I gave, or any other reasonable example
of a fixed wing airplane in horizontal flight.

...

That's the shortcut. Where does the Bernoulli effect come from - on a
molecular level? That's what I'm addressing. The Bernoulli effect is a
shortcut for doing the calculation in bulk (where it makes the most
sense if you want a numerical answer) but it all comes from molecular
collisions.

I agreed quite some time ago that the theoretical basis for
macroscopic gas laws is to be found in statistical mechanics.

On a macroscopic level, the vertical component of momentum of the
wing is zero. Therefor on a macroscopic level, the sum of the
momenta transferred to the air molecules, integrated over all of
the air molecules must also be zero by Newton's third law.

Right?

For an airplane in straight level flight there is no net momentum
transfer in the vertical direction, between the air and the airplane,
just like there is no net vertical force acting on the airplane.

--

FF

If so, then during the time it is in freefall, it acquires a downward
velocity. Small, no doubt, but nonzero.

Sometimes it does, sometimes it does not. I'll allow as the vertical
component of velocity decreases during that time, for a positive up
coordinate system and a plane in (macroscopic) level flight.

Ok. (I was sloppy - it doesn't "acquire a downward velocity", it really
"endures a downward acceleration", which depending on the initial
vertical velocity may or may not end up with the plane going downward.)
So we are saying the same thing here.

Do you agree that in each collision momentum is transferred to the
air molecule that is equal and opposite to the momentum transferred
to the wing?

Yes I do. This is what I call "throwing the air down". That downward
momentum will remain with the air (dissipated across many other
molecules as it keeps colliding, but never disappearing) until it is
transferred to the earth, which has been accelerating upwards in the
same fashion.

I agreed quite some time ago that the theoretical basis for
macroscopic gas laws is to be found in statistical mechanics.

Ok.

On a macroscopic level, the vertical component of momentum of the
wing is zero.

Yes.

Therefor on a macroscopic level, the sum of the
momenta transferred to the air molecules, integrated over all of
the air molecules must also be zero by Newton's third law.

Right?

Only in a nonaccelerated frame. We are dealing with an accelerated
frame. Consider a rocketship hovering over the moon. The (macroscopic)
vertical component of its momentum is zero also. However it has to
continually throw down rocket exhaust to stay there. So, without
looking at the rest of the picture, your conclusion about momentum is
flawed.

In the case of the wing, the momentum is transferred a few times... once
when the wing hits the air molecule (throwing the air down), again when
that molecule hits the earth and bounces back (throwing the earth away
from the wing), and then again when that air molecule (or its proxy)
hits the wing on the way back up.

Think about a person sitting on a stool. No momentum transfer (or so it
would seem). But then think about a person supporting himself by
dribbling a basketball. There is a lot of momentum transfer, but no
=net= change. The reason there is no net change is that the basketball
keeps pushing the earth away too.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.